# If y = $\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}$, find $\rm \frac{{dy}}{{dx}}$

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If y = $\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}$, find $\rm \frac{{dy}}{{dx}}$
1. $\rm\frac{1}{2}\left( {secx.cotx + cosecx.tanx} \right)$
2. $\rm \frac{1}{2}\left( {secx.cotx - cosecx.tanx} \right)$
3. $\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)$
4. $\rm \frac{1}{2}\left( {secx.tanx - cosecx.cotx} \right)$

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Correct Answer - Option 3 : $\rm \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)$

Concept used:

Trigonometry formula

sin 2x = 2sin x cos x

Calculation:

y = $\rm {\rm{\;}}\frac{{\left( {sinx - cosx} \right)}}{{sin2x}}$

y = $\rm \frac{{\left( {sinx - cosx} \right)}}{{2.sinx.cosx}}$

⇒ y = $\rm \frac{{sinx}}{{2.sinx.cosx}} - \;\frac{{cosx}}{{2.sinx.cosx}}$

⇒ y = $\rm \left( {\frac{1}{{2cosx}}} \right)\; - \left( {\frac{1}{{2sinx}}} \right)$

⇒ y = $\frac{1}{2}$ (secx - cosecx)

Differentiate both sides, we get

⇒ $\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left[ {\frac{{d\left( {secx} \right)}}{{dx}} - \frac{{d\left( {cosecx} \right)}}{{dx}}} \right]$

⇒ $\rm \frac{{dy}}{{dx}} = \frac{1}{2}\left( {secx.tanx + cosecx.cotx} \right)$